Ch 6 data and computer communicationwilliam stallings

DIGITAL DATA COMMUNICATION
CHAPTER
6
TECHNIQUES
6.1 Asynchronous and Synchronous Transmission

6.2 Types of Errors

6.3 Error Detection

6.4 Error Correction

6.5 Line Configurations

6.6 Recommended Reading

6.7 Key Terms, Review Questions, and Problems

180

A conversation forms a two-way communication link; there is a measure of symme-
try between the two parties, and messages pass to and fro. There is a continual stim-
ulus-response, cyclic action; remarks call up other remarks, and the behavior of the
two individuals becomes concerted, co-operative, and directed toward some goal.
This is true communication.

—On Human Communication, Colin Cherry

KEY POINTS
• The transmission of a stream of bits from one device to another
across a transmission link involves a great deal of cooperation and
agreement between the two sides. One of the most fundamental
requirements is synchronization. The receiver must know the rate at
which bits are being received so that it can sample the line at appro-
priate intervals to determine the value of each received bit. Two tech-
niques are in common use for this purpose. In asynchronous
transmission, each character of data is treated independently. Each
character begins with a start bit that alerts the receiver that a charac-
ter is arriving. The receiver samples each bit in the character and then
looks for the beginning of the next character. This technique would
not work well for long blocks of data because the receiver’s clock
might eventually drift out of synchronization with the transmitter’s
clock. However, sending data in large blocks is more efficient than
sending data one character at a time. For large blocks, synchronous
transmission is used. Each block of data is formatted as a frame that
includes a starting and an ending flag. Some form of synchronization,
such as the use of Manchester encoding, is employed.
• Error detection is performed by calculating an error-detecting code
that is a function of the bits being transmitted. The code is appended
to the transmitted bits. The receiver calculates the code based on the
incoming bits and compares it to the incoming code to check for
errors.
• Error correction operates in a fashion similar to error detection but is
capable of correcting certain errors in a transmitted bit stream.

The preceding three chapters have been concerned primarily with the
attributes of data transmission, such as the characteristics of data signals
and transmission media, the encoding of signals, and transmission perform-
ance. In this chapter, we shift our emphasis from data transmission to data
communications.
181

182 CHAPTER 6 / DIGITAL DATA COMMUNICATION TECHNIQUES

For two devices linked by a transmission medium to exchange data, a
high degree of cooperation is required. Typically, data are transmitted one bit
at a time over the medium. The timing (rate, duration, spacing) of these bits
must be the same for transmitter and receiver. Two common techniques for
controlling this timing—asynchronous and synchronous—are explored in
Section 6.1. Next, we look at the problem of bit errors. As we have seen, data
transmission is not an error-free process, and some means of accounting for
these errors is needed. After a brief discussion of the distinction between sin-
gle-bit errors and burst errors, the chapter turns to two approaches to dealing
with errors: error detection and error correction.
Next, the chapter provides an overview of the types of line configurations
in common use.To supplement the material in this chapter,Appendix G looks at
the physical interface between data transmitting devices and the transmission
line. Typically, digital data devices do not attach to and signal across the medium
directly. Instead, this process is mediated through a standardized interface that
provides considerable control over the interaction between the transmitting/
receiving devices and the transmission line.

6.1 ASYNCHRONOUS AND SYNCHRONOUS TRANSMISSION

In this book, we are primarily concerned with serial transmission of data; that is,
data are transferred over a single signal path rather than a parallel set of lines, as is
common with I/O devices and internal computer signal paths. With serial transmis-
sion, signaling elements are sent down the line one at a time. Each signaling element
may be
• Less than one bit: This is the case, for example, with Manchester coding.
• One bit: NRZ-L and FSK are digital and analog examples, respectively.
• More than one bit: QPSK is an example.
For simplicity in the following discussion, we assume one bit per signal-
ing element unless otherwise stated. The discussion is not materially affected by
this simplification.
Recall from Figure 3.16 that the reception of digital data involves sampling
the incoming signal once per bit time to determine the binary value. One of the
difficulties encountered in such a process is that various transmission impairments
will corrupt the signal so that occasional errors will occur. This problem is com-
pounded by a timing difficulty: In order for the receiver to sample the incoming
bits properly, it must know the arrival time and duration of each bit that it
receives.
Suppose that the sender simply transmits a stream of data bits. The sender has
a clock that governs the timing of the transmitted bits. For example, if data are to be
transmitted at one million bits per second (1 Mbps), then one bit will be transmitted
every 1/106 = 1 microsecond 1ms2, as measured by the sender’s clock. Typically,
the receiver will attempt to sample the medium at the center of each bit time. The
receiver will time its samples at intervals of one bit time. In our example, the

6.1 / ASYNCHRONOUS AND SYNCHRONOUS TRANSMISSION 183
sampling would occur once every 1 ms. If the receiver times its samples based on its
own clock, then there will be a problem if the transmitter’s and receiver’s clocks are
not precisely aligned. If there is a drift of 1% (the receiver’s clock is 1% faster or
slower than the transmitter’s clock), then the first sampling will be 0.01 of a bit time
10.01 ms2 away from the center of the bit (center of bit is 0.5 ms from beginning and
end of bit). After 50 or more samples, the receiver may be in error because it is sam-
pling in the wrong bit time 150 * .01 = 0.5 ms2. For smaller timing differences, the
error would occur later, but eventually the receiver will be out of step with the trans-
mitter if the transmitter sends a sufficiently long stream of bits and if no steps are
taken to synchronize the transmitter and receiver.

Asynchronous Transmission
Two approaches are common for achieving the desired synchronization. The first is
called, oddly enough, asynchronous transmission. The strategy with this scheme is to
avoid the timing problem by not sending long, uninterrupted streams of bits.
Instead, data are transmitted one character at a time, where each character is five to
eight bits in length.1 Timing or synchronization must only be maintained within each
character; the receiver has the opportunity to resynchronize at the beginning of
each new character.
Figure 6.1 illustrates this technique. When no character is being transmitted,
the line between transmitter and receiver is in an idle state. The definition of idle is
equivalent to the signaling element for binary 1. Thus, for NRZ-L signaling (see
Figure 5.2), which is common for asynchronous transmission, idle would be the pres-
ence of a negative voltage on the line. The beginning of a character is signaled by a
start bit with a value of binary 0. This is followed by the 5 to 8 bits that actually make
up the character. The bits of the character are transmitted beginning with the least
significant bit. For example, for IRA characters, the data bits are usually followed by
a parity bit, which therefore is in the most significant bit position. The parity bit is set
by the transmitter such that the total number of ones in the character, including the
parity bit, is even (even parity) or odd (odd parity), depending on the convention
being used. The receiver uses this bit for error detection, as discussed in Section 6.3.
The final element is a stop element, which is a binary 1. A minimum length for the
stop element is specified, and this is usually 1, 1.5, or 2 times the duration of an ordi-
nary bit. No maximum value is specified. Because the stop element is the same as
the idle state, the transmitter will continue to transmit the stop element until it is
ready to send the next character.
The timing requirements for this scheme are modest. For example, IRA char-
acters are typically sent as 8-bit units, including the parity bit. If the receiver is 5%
slower or faster than the transmitter, the sampling of the eighth character bit will be
displaced by 45% and still be correctly sampled.

1
The number of bits that comprise a character depends on the code used. We have already described one
common example, the IRA code, which uses seven bits per character. Another common code is the
Extended Binary Coded Decimal Interchange Code (EBCDIC), which is an 8-bit character code used on
IBM mainframes.


Idle state Odd or even Remain idle or
of line parity or unused next start bit
5 to 8 data bits 1 to 2 bit times

0
Start P Stop
bit bit element
1
(a) Character format

Unpredictable time interval
between characters
Stop Stop
Start element Start element
bit bit

1 1 0 1 0 0 0 1 0 1 1 0 1 1 0 0
(b) 8-bit asynchronous character stream

50 150 250 350 450 550 650 750 850 Transmitter timing (␮s)

0
Start Stop
bit 1 2 3 4 5 6 7 8 element
1

47 141 235 329 423 517 611 705 799 Receiver timing (␮s)
(c) Effect of timing error

Figure 6.1 Asynchronous Transmission

EXAMPLE 6.1 Figure 6.1c shows the effects of a timing error of sufficient
magnitude to cause an error in reception. In this example we assume a data rate
of 10,000 bits per second (10 kbps); therefore, each bit is of 0.1 millisecond (ms),
or 100 ms, duration. Assume that the receiver is fast by 6%, or 6 ms per bit time.
Thus, the receiver samples the incoming character every 94 ms (based on the
transmitter’s clock). As can be seen, the last sample is erroneous.

An error such as just described actually results in two errors. First, the last
sampled bit is incorrectly received. Second, the bit count may now be out of align-
ment. If bit 7 is a 1 and bit 8 is a 0, bit 8 could be mistaken for a start bit. This condi-
tion is termed a framing error, as the character plus start bit and stop element are
sometimes referred to as a frame. A framing error can also occur if some noise con-
dition causes the false appearance of a start bit during the idle state.
Asynchronous transmission is simple and cheap but requires an overhead of
two to three bits per character. For example, for an 8-bit character with no parity
bit, using a 1-bit-long stop element, two out of every ten bits convey no informa-
tion but are there merely for synchronization; thus the overhead is 20%. Of

6.1 / ASYNCHRONOUS AND SYNCHRONOUS TRANSMISSION 185
course, the percentage overhead could be reduced by sending larger blocks of bits
between the start bit and stop element. However, as Figure 6.1c indicates, the
larger the block of bits, the greater the cumulative timing error. To achieve greater
efficiency, a different form of synchronization, known as synchronous transmis-
sion, is used.

Synchronous Transmission
With synchronous transmission, a block of bits is transmitted in a steady stream
without start and stop codes. The block may be many bits in length. To prevent tim-
ing drift between transmitter and receiver, their clocks must somehow be synchro-
nized. One possibility is to provide a separate clock line between transmitter and
receiver. One side (transmitter or receiver) pulses the line regularly with one short
pulse per bit time. The other side uses these regular pulses as a clock. This technique
works well over short distances, but over longer distances the clock pulses are sub-
ject to the same impairments as the data signal, and timing errors can occur. The
other alternative is to embed the clocking information in the data signal. For digital
signals, this can be accomplished with Manchester or differential Manchester encod-
ing. For analog signals, a number of techniques can be used; for example, the carrier
frequency itself can be used to synchronize the receiver based on the phase of the
carrier.
With synchronous transmission, there is another level of synchronization
required, to allow the receiver to determine the beginning and end of a block of data.
To achieve this, each block begins with a preamble bit pattern and generally ends
with a postamble bit pattern. In addition, other bits are added to the block that con-
vey control information used in the data link control procedures discussed in
Chapter 7. The data plus preamble, postamble, and control information are called a
frame. The exact format of the frame depends on which data link control procedure
is being used.
Figure 6.2 shows, in general terms, a typical frame format for synchronous
transmission. Typically, the frame starts with a preamble called a flag, which is 8 bits
long. The same flag is used as a postamble. The receiver looks for the occurrence of
the flag pattern to signal the start of a frame. This is followed by some number of
control fields (containing data link control protocol information), then a data field
(variable length for most protocols), more control fields, and finally the flag is
repeated.
For sizable blocks of data, synchronous transmission is far more efficient than
asynchronous. Asynchronous transmission requires 20% or more overhead. The
control information, preamble, and postamble in synchronous transmission are typ-
ically less than 100 bits.

8-bit Control Control 8-bit
Data field
flag fields fields flag

Figure 6.2 Synchronous Frame Format


EXAMPLE 6.2 One of the more common schemes, HDLC (described in Chapter
7), contains 48 bits of control, preamble, and postamble. Thus, for a 1000-character
block of data, each frame consists of 48 bits of overhead and 1000 * 8 = 8,000 bits
of data, for a percentage overhead of only 48/8048 * 100% = 0.6%.

6.2 TYPES OF ERRORS

In digital transmission systems, an error occurs when a bit is altered between trans-
mission and reception; that is, a binary 1 is transmitted and a binary 0 is received, or
a binary 0 is transmitted and a binary 1 is received. Two general types of errors can
occur: single-bit errors and burst errors. A single-bit error is an isolated error condi-
tion that alters one bit but does not affect nearby bits. A burst error of length B is a
contiguous sequence of B bits in which the first and last bits and any number of
intermediate bits are received in error. More precisely, IEEE Std 100 and ITU-T
Recommendation Q.9 both define an error burst as follows:

Error burst: A group of bits in which two successive erroneous bits are
always separated by less than a given number x of correct bits. The last erro-
neous bit in the burst and the first erroneous bit in the following burst are
accordingly separated by x correct bits or more.

Thus, in an error burst, there is a cluster of bits in which a number of errors
occur, although not necessarily all of the bits in the cluster suffer an error.
A single-bit error can occur in the presence of white noise, when a slight ran-
dom deterioration of the signal-to-noise ratio is sufficient to confuse the receiver’s
decision of a single bit. Burst errors are more common and more difficult to deal
with. Burst errors can be caused by impulse noise, which was described in Chapter 3.
Another cause is fading in a mobile wireless environment; fading is described in
Chapter 14.
Note that the effects of burst errors are greater at higher data rates.

EXAMPLE 6.3 An impulse noise event or a fading event of 1 ms occurs. At a
data rate of 10 Mbps, there is a resulting error burst of 10 bits. At a data rate of
100 Mbps, there is an error burst of 100 bits.

6.3 ERROR DETECTION

Regardless of the design of the transmission system, there will be errors, resulting
in the change of one or more bits in a transmitted frame. In what follows, we

6.3 / ERROR DETECTION 187
assume that data are transmitted as one or more contiguous sequences of bits,
called frames. We define these probabilities with respect to errors in transmitted
frames:
Pb: Probability that a bit is received in error; also known as the bit error rate
(BER)
P1: Probability that a frame arrives with no bit errors
P2: Probability that, with an error-detecting algorithm in use, a frame arrives with
one or more undetected errors
P3: Probability that, with an error-detecting algorithm in use, a frame arrives
with one or more detected bit errors but no undetected bit errors
First consider the case in which no means are taken to detect errors. Then the
probability of detected errors 1P32 is zero. To express the remaining probabilities,
assume the probability that any bit is in error 1Pb2 is constant and independent for
each bit. Then we have
P1 = 11 - Pb2F
P2 = 1 - P1
where F is the number of bits per frame. In words, the probability that a frame
arrives with no bit errors decreases when the probability of a single bit error
increases, as you would expect. Also, the probability that a frame arrives with no bit
errors decreases with increasing frame length; the longer the frame, the more bits it
has and the higher the probability that one of these is in error.

EXAMPLE 6.4 A defined objective for ISDN (integrated services digital net-
work) connections is that the BER on a 64-kbps channel should be less than 10-6
on at least 90% of observed 1-minute intervals. Suppose now that we have the
rather modest user requirement that on average one frame with an undetected
bit error should occur per day on a continuously used 64-kbps channel, and let
us assume a frame length of 1000 bits. The number of frames that can be trans-
mitted in a day comes out to 5.529 * 106, which yields a desired frame error
rate of P2 = 1/15.529 * 1062 = 0.18 * 10-6. But if we assume a value of Pb of
10-6, then P1 = 10.99999921000 = 0.999 and therefore P2 = 10 -3, which is about
three orders of magnitude too large to meet our requirement.

This is the kind of result that motivates the use of error-detecting techniques.All
of these techniques operate on the following principle (Figure 6.3). For a given frame
of bits, additional bits that constitute an error-detecting code are added by the trans-
mitter. This code is calculated as a function of the other transmitted bits. Typically, for
a data block of k bits, the error-detecting algorithm yields an error-detecting code of
n - k bits, where 1n - k2 6 k. The error-detecting code, also referred to as the
check bits, is appended to the data block to produce a frame of n bits, which is then


k bits

Data Data'

E ‫ ؍‬f(data) E' ‫ ؍‬f(data') COMPARE

Receiver
Data

n ؊ k bits E, E' ‫ ؍‬error-detecting codes
f ‫ ؍‬error-detecting code function
n bits

Transmitter

Figure 6.3 Error Detection Process

transmitted. The receiver separates the incoming frame into the k bits of data and
1n - k2 bits of the error-detecting code.The receiver performs the same error-detect-
ing calculation on the data bits and compares this value with the value of the incoming
error-detecting code. A detected error occurs if and only if there is a mismatch. Thus
P3 is the probability that a frame contains errors and that the error-detecting scheme
will detect that fact. P2 is known as the residual error rate and is the probability that an
error will be undetected despite the use of an error-detecting scheme.

Parity Check
The simplest error-detecting scheme is to append a parity bit to the end of a block of
data. A typical example is character transmission, in which a parity bit is attached to
each 7-bit IRA character. The value of this bit is selected so that the character has an
even number of 1s (even parity) or an odd number of 1s (odd parity).

EXAMPLE 6.5 If the transmitter is transmitting an IRA character G (1110001)
and using odd parity, it will append a 1 and transmit 11110001.2 The receiver exam-
ines the received character and, if the total number of 1s is odd, assumes that no
error has occurred. If one bit (or any odd number of bits) is erroneously inverted
during transmission (for example, 11100001), then the receiver will detect an error.

2
Recall from our discussion in Section 5.1 that the least significant bit of a character is transmitted first
and that the parity bit is the most significant bit.

Note, however, that if two (or any even number) of bits are inverted due to
error, an undetected error occurs. Typically, even parity is used for synchronous
transmission and odd parity for asynchronous transmission.
The use of the parity bit is not foolproof, as noise impulses are often long
enough to destroy more than one bit, particularly at high data rates.

Cyclic Redundancy Check (CRC)
One of the most common, and one of the most powerful, error-detecting codes is the
cyclic redundancy check (CRC), which can be described as follows. Given a k-bit
block of bits, or message, the transmitter generates an 1n - k2-bit sequence, known
as a frame check sequence (FCS), such that the resulting frame, consisting of n bits,
is exactly divisible by some predetermined number. The receiver then divides the
incoming frame by that number and, if there is no remainder, assumes there was no
error.3
To clarify this, we present the procedure in three equivalent ways: modulo 2
arithmetic, polynomials, and digital logic.

Modulo 2 Arithmetic Modulo 2 arithmetic uses binary addition with no carries,
which is just the exclusive-OR (XOR) operation. Binary subtraction with no carries
is also interpreted as the XOR operation: For example,

1111 1111 11001
ϩ1010
_______ Ϫ0101
_______ ϫ 11
_______
0101 1010 11001
11001
_______
101011

Now define

T = n-bit frame to be transmitted
D = k-bit block of data, or message, the first k bits of T
F = 1n - k2-bit FCS, the last 1n - k2 bits of T
P = pattern of n - k + 1 bits; this is the predetermined divisor

We would like T/P to have no remainder. It should be clear that

T = 2 n - kD + F

That is, by multiplying D by 2 n - k, we have in effect shifted it to the left by n - k bits
and padded out the result with zeroes. Adding F yields the concatenation of D and

3
This procedure is slightly different from that of Figure 6.3. As shall be seen, the CRC process could be
implemented as follows. The receiver could perform a division operation on the incoming k data bits and
compare the result to the incoming 1n - k2 check bits.


F, which is T. We want T to be exactly divisible by P. Suppose that we divide 2 n - kD
by P:

2 n - kD R
= Q + (6.1)
P P

There is a quotient and a remainder. Because division is modulo 2, the remainder is
always at least one bit shorter than the divisor. We will use this remainder as our
FCS. Then

T = 2 n - kD + R (6.2)

Does this R satisfy our condition that T/P have no remainder? To see that it does,
consider

T 2 n - kD + R 2 n - kD R
= = +
P P P P

Substituting Equation (6.1), we have

T R R
= Q + +
P P P

However, any binary number added to itself modulo 2 yields zero. Thus

T R + R
= Q + = Q
P P

There is no remainder, and therefore T is exactly divisible by P. Thus, the FCS is eas-
ily generated: Simply divide 2 n - kD by P and use the 1n - k2-bit remainder as the
FCS. On reception, the receiver will divide T by P and will get no remainder if there
have been no errors.

EXAMPLE 6.6

1. Given

Message D = 1010001101 110 bits2
Pattern P = 110101 16 bits2
FCS R = to be calculated 15 bits2
Thus, n = 15, k = 10, and 1n - k2 = 5.
2. The message is multiplied by 2 5, yielding 101000110100000.
3. This product is divided by P:


1 1 0 1 0 1 0 1 1 0 Q
P 1 1 0 1 0 1 1 0 1 0 0 0 1 1 0 1 0 0 0 0 0 2n–kD
1 1 0 1 0 1
1 1 1 0 1 1
1 1 0 1 0 1
1 1 1 0 1 0
1 1 0 1 0 1
1 1 1 1 1 0
1 1 0 1 0 1
1 0 1 1 0 0
1 1 0 1 0 1
1 1 0 0 1 0
1 1 0 1 0 1
0 1 1 1 0 R

4. The remainder is added to 2 5D to give T = 101000110101110, which is trans-
mitted.
5. If there are no errors, the receiver receives T intact. The received frame is
divided by P:

1 1 0 1 0 1 0 1 1 0 Q
P 1 1 0 1 0 1 1 0 1 0 0 0 1 1 0 1 0 1 1 1 0 T
1 1 0 1 0 1
1 1 1 0 1 1
1 1 0 1 0 1
1 1 1 0 1 0
1 1 0 1 0 1
1 1 1 1 1 0
1 1 0 1 0 1
1 0 1 1 1 1
1 1 0 1 0 1
1 1 0 1 0 1
1 1 0 1 0 1
0 R

Because there is no remainder, it is assumed that there have been no errors.

The pattern P is chosen to be one bit longer than the desired FCS, and the
exact bit pattern chosen depends on the type of errors expected. At minimum, both
the high- and low-order bits of P must be 1.
There is a concise method for specifying the occurrence of one or more errors.
An error results in the reversal of a bit. This is equivalent to taking the XOR of the


bit and 1 (modulo 2 addition of 1 to the bit): 0 + 1 = 1; 1 + 1 = 0. Thus, the errors
in an n-bit frame can be represented by an n-bit field with 1s in each error position.
The resulting frame Tr can be expressed as

Tr = T ᮍ E

where
T = transmitted frame
E = error pattern with 1s in positions where errors occur
Tr = received frame
ᮍ = bitwise exclusive-OR1XOR2
If there is an error 1E Z 02, the receiver will fail to detect the error if and only if Tr
is divisible by P, which is equivalent to E divisible by P. Intuitively, this seems an
unlikely occurrence.

Polynomials A second way of viewing the CRC process is to express all values as
polynomials in a dummy variable X, with binary coefficients. The coefficients
correspond to the bits in the binary number. Thus, for D = 110011, we have
D1X2 = X5 + X4 + X + 1, and for P = 11001, we have P1X2 = X4 + X3 + 1.
Arithmetic operations are again modulo 2. The CRC process can now be described
as
Xn - kD1X2 R1X2
= Q1X2 +
P1X2 P1X2
n-k
T1X2 = X D1X2 + R1X2
Compare these equations with Equations (6.1) and (6.2).

EXAMPLE 6.7 Using the preceding example, for D = 1010001101, we have
D1X2 = X9 + X7 + X3 + X2 + 1, and for P = 110101, we have
P1X2 = X5 + X4 + X2 + 1. We should end up with R = 01110, which corre-
sponds to R1X2 = X3 + X2 + X. Figure 6.4 shows the polynomial division that
corresponds to the binary division in the preceding example.

An error E(X) will only be undetectable if it is divisible by P(X). It can be
shown [PETE 61, RAMA88] that all of the following errors are not divisible by a
suitably chosen P(X) and hence are detectable:
• All single-bit errors, if P(X) has more than one nonzero term
• All double-bit errors, as long as P(X) is a special type of polynomial, called a
primitive polynomial, with maximum exponent L, and the frame length is less
than 2 L - 1.


X9 ؉ X8 ؉ X6 ؉ X4 ؉ X2 ؉ X Q(X)
P(X) X ؉X ؉X ؉1
5 4 2
X 14
X 12
X ؉X ؉
8 7
X 5 X5D(X)
X ؉X ؉
14 13 X ؉
11
X 9

X13 ؉ X12 ؉ X11 ؉ X9 ؉ X8
X13 ؉ X12 ؉ X10 ؉ X8
X ؉X ؉X ؉
11 10 9
X7
X ؉X ؉
11 10
X ؉
8
X6
X ؉ X ؉ X ؉ X6 ؉ X5
9 8 7

X9 ؉ X8 ؉ X6 ؉ X4
X7 ؉ X ؉ X4
5

X7 ؉ X6 ؉ X4 ؉ X2
X ؉X ؉
6 5
X2
X6 ؉ X5 ؉ X ؉
3
X
X3 ؉ X2 ؉ X R(X)

Figure 6.4 Example of Polynomial Division

• Any odd number of errors, as long as P(X) contains a factor 1X + 12
• Any burst error for which the length of the burst is less than or equal to n - k;
that is, less than or equal to the length of the FCS
• A fraction of error bursts of length n - k + 1; the fraction equals
1 - 2 -1n - k - 12
• A fraction of error bursts of length greater than n - k + 1; the fraction
equals 1 - 2 -1n - k2

In addition, it can be shown that if all error patterns are considered equally
likely, then for a burst error of length r + 1, the probability of an undetected error
(E(X) is divisible by P(X)) is 1/2 r - 1, and for a longer burst, the probability is 1/2 r,
where r is the length of the FCS.
Four versions of P(X) are widely used:

CRC-12 = X12 + X11 + X3 + X2 + X + 1
CRC-16 = X16 + X15 + X2 + 1
CRC-CCITT = X16 + X12 + X5 + 1
CRC-32 = X32 + X26 + X23 + X22 + X16 + X12 + X11
+ X10 + X8 + X7 + X5 + X4 + X2 + X + 1

The CRC-12 system is used for transmission of streams of 6-bit characters and gen-
erates a 12-bit FCS. Both CRC-16 and CRC-CCITT are popular for 8-bit characters,
in the United States and Europe, respectively, and both result in a 16-bit FCS. This
would seem adequate for most applications, although CRC-32 is specified as an
option in some point-to-point synchronous transmission standards and is used in
IEEE 802 LAN standards.


Digital Logic The CRC process can be represented by, and indeed imple-
mented as, a dividing circuit consisting of XOR gates and a shift register. The shift
register is a string of 1-bit storage devices. Each device has an output line, which
indicates the value currently stored, and an input line. At discrete time instants,
known as clock times, the value in the storage device is replaced by the value indi-
cated by its input line. The entire register is clocked simultaneously, causing a 1-bit
shift along the entire register. The circuit is implemented as follows:
1. The register contains n - k bits, equal to the length of the FCS.
2. There are up to n - k XOR gates.
3. The presence or absence of a gate corresponds to the presence or absence of
a term in the divisor polynomial, P(X), excluding the terms 1 and Xn - k.

EXAMPLE 6.8 The architecture of a CRC circuit is best explained by first con-
sidering an example, which is illustrated in Figure 6.5. In this example, we use

Data D = 1010001101; D1X2 = X9 + X7 + X3 + X2 + 1
Divisor P = 110101; P1X2 = X5 + X4 + X2 + 1

which were used earlier in the discussion.
Figure 6.5a shows the shift register implementation. The process begins with
the shift register cleared (all zeros). The message, or dividend, is then entered,
one bit at a time, starting with the most significant bit. Figure 6.5b is a table that
shows the step-by-step operation as the input is applied one bit at a time. Each
row of the table shows the values currently stored in the five shift-register ele-
ments. In addition, the row shows the values that appear at the outputs of the
three XOR circuits. Finally, the row shows the value of the next input bit, which is
available for the operation of the next step.
Note that the XOR operation affects C4 , C2 , and C0 on the next shift. This is
identical to the binary long division process illustrated earlier. The process con-
tinues through all the bits of the message. To produce the proper output, two
switches are used. The input data bits are fed in with both switches in the A posi-
tion. As a result, for the first 10 steps, the input bits are fed into the shift register
and also used as output bits. After the last data bit is processed, the shift register
contains the remainder (FCS) (shown shaded). As soon as the last data bit is pro-
vided to the shift register, both switches are set to the B position. This has two
effects: (1) All of the XOR gates become simple pass-throughs; no bits are
changed, and (2) as the shifting process continues, the 5 CRC bits are output.
At the receiver, the same logic is used. As each bit of M arrives, it is inserted
into the shift register. If there have been no errors, the shift register should con-
tain the bit pattern for R at the conclusion of M. The transmitted bits of R now
begin to arrive, and the effect is to zero out the register so that, at the conclusion
of reception, the register contains all 0s.


Output
(15 bits)

Switch 1
A B
Input
(10 bits)
C4 ᮍ C3 C2 ᮍ C1 C0
Switch
2 A
ᮍ
B
ϭ 1-bit shift register ᮍ ϭ Exclusive-OR circuit

(a) Shift-register implementation

C4 C3 C2 C1 C0 C4ᮍC3ᮍI C4ᮍC1ᮍI C4ᮍI I ϭ input
Initial 0 0 0 0 0 1 1 1 1
Step 1 1 0 1 0 1 1 1 1 0
Step 2 1 1 1 1 1 1 1 0 1
Step 3 1 1 1 1 0 0 0 1 0
Step 4 0 1 0 0 1 1 0 0 0 Message to
Step 5 1 0 0 1 0 1 0 1 0 be sent
Step 6 1 0 0 0 1 0 0 0 1
Step 7 0 0 0 1 0 1 0 1 1
Step 8 1 0 0 0 1 1 1 1 0
Step 9 1 0 1 1 1 0 1 0 1
Step 10 0 1 1 1 0
(b) Example with input of 1010001101

Figure 6.5 Circuit with Shift Registers for Dividing by the Polynomial
X5 + X4 + X2 + 1

Output
(n bits)

Switch 1
A B
Input
(k bits)
CnϪkϪ1 ᮍ CnϪkϪ2 ᮍ ᮍ C1 ᮍ C0

؋ AnϪkϪ1 ؋ AnϪkϪ2 ؋ A2 ؋ A1
Switch
2 A
ᮍ
B
Figure 6.6 General CRC Architecture to Implement Divisor 11 + A 1X + A 2X2 + Á +
A n - 1X n - k - 1 + X n - k2


Figure 6.6 indicates the general architecture of the shift register implementa-
n-k
tion of a CRC for the polynomial P1X2 = g i = 0 A iXi, where A 0 = A n - k = 1 and
4
all other A i equal either 0 or 1.

6.4 ERROR CORRECTION

Error detection is a useful technique, found in data link control protocols, such as
HDLC, and in transport protocols, such as TCP. However, correction of errors
using an error-detecting code, requires that block of data be retransmitted, as
explained in Chapter 7. For wireless applications this approach is inadequate for
two reasons.
1. The bit error rate on a wireless link can be quite high, which would result in a
large number of retransmissions.
2. In some cases, especially satellite links, the propagation delay is very long com-
pared to the transmission time of a single frame. The result is a very inefficient
system. As is discussed in Chapter 7, the common approach to retransmission
is to retransmit the frame in error plus all subsequent frames. With a long data
link, an error in a single frame necessitates retransmitting many frames.
Instead, it would be desirable to enable the receiver to correct errors in an
incoming transmission on the basis of the bits in that transmission. Figure 6.7 shows
in general how this is done. On the transmission end, each k-bit block of data is

k bits
Codeword
Data

FEC
FEC
decoder
Detectable but not
correctable error

encoder
No error or
correctable
error

Codeword

n bits
Data

Transmitter Receiver

Figure 6.7 Error Correction Process

4
It is common for the CRC register to be shown shifting to the right, which is the reverse of the analogy
to binary division. Because binary numbers are usually shown with the most significant bit on the left, a
left-shifting register, as is used here, is more appropriate.

6.4 / ERROR CORRECTION 197
mapped into an n-bit block 1n 7 k2 called a codeword, using an FEC (forward
error correction) encoder. The codeword is then transmitted. During transmission,
the signal is subject to impairments, which may produce bit errors in the signal. At
the receiver, the incoming signal is demodulated to produce a bit string that is simi-
lar to the original codeword but may contain errors. This block is passed through an
FEC decoder, with one of four possible outcomes:
1. If there are no bit errors, the input to the FEC decoder is identical to the
original codeword, and the decoder produces the original data block as
output.
2. For certain error patterns, it is possible for the decoder to detect and correct
those errors. Thus, even though the incoming data block differs from the trans-
mitted codeword, the FEC decoder is able to map this block into the original
data block.
3. For certain error patterns, the decoder can detect but not correct the errors. In
this case, the decode simply reports an uncorrectable error.
4. For certain, typically rare, error patterns, the decoder does not detect that any
errors have occurred and maps the incoming n-bit data block into a k-bit block
that differs from the original k-bit block.
How is it possible for the decoder to correct bit errors? In essence, error cor-
rection works by adding redundancy to the transmitted message. The redundancy
makes it possible for the receiver to deduce what the original message was, even in
the face of a certain level of error rate. In this section we look at a widely used form
of error-correcting code known as a block error-correcting code. Our discussion
only deals with basic principles; a discussion of specific error-correcting codes is
beyond our scope.
Before proceeding, we note that in many cases, the error-correcting code follows
the same general layout as shown for error-detecting codes in Figure 6.3. That is, the
FEC algorithm takes as input a k-bit block and adds 1n - k2 check bits to that block
to produce an n-bit block; all of the bits in the original k-bit block show up in the n-bit
block. For some FEC algorithms, the FEC algorithm maps the k-bit input into an n-bit
codeword in such a way that the original k bits do not appear in the codeword.

Block Code Principles
To begin, we define a term that shall be of use to us. The Hamming distance d1v1 , v22
between two n-bit binary sequences v1 and v2 is the number of bits in which v1 and
v2 disagree. For example, if
v1 = 011011, v2 = 110001
then
d1v1 , v22 = 3
Now let us consider the block code technique for error correction. Suppose we
wish to transmit blocks of data of length k bits. Instead of transmitting each block as
k bits, we map each k-bit sequence into a unique n-bit codeword.


EXAMPLE 6.9

For k = 2 and n = 5, we can make the following assignment:
Data Block Codeword
00 00000
01 00111
10 11001
11 11110
Now, suppose that a codeword block is received with the bit pattern 00100. This is
not a valid codeword, and so the receiver has detected an error. Can the error be
corrected? We cannot be sure which data block was sent because 1, 2, 3, 4, or even
all 5 of the bits that were transmitted may have been corrupted by noise. How-
ever, notice that it would require only a single bit change to transform the valid
codeword 00000 into 00100. It would take two bit changes to transform 00111 to
00100, three bit changes to transform 11110 to 00100, and it would take four bit
changes to transform 11001 into 00100. Thus, we can deduce that the most likely
codeword that was sent was 00000 and that therefore the desired data block is 00.
This is error correction. In terms of Hamming distances, we have

d100000, 001002 = 1; d100111, 001002 = 2;
d111001, 001002 = 4; d111110, 001002 = 3

So the rule we would like to impose is that if an invalid codeword is received,
then the valid codeword that is closest to it (minimum distance) is selected. This
will only work if there is a unique valid codeword at a minimum distance from
each invalid codeword.
For our example, it is not true that for every invalid codeword there is one
and only one valid codeword at a minimum distance. There are 2 5 = 32 possible
codewords of which 4 are valid, leaving 28 invalid codewords. For the invalid
codewords, we have the following:
Invalid Minimum Valid Invalid Minimum Valid
Codeword Distance Codeword Codeword Distance Codeword

00001 1 00000 10000 1 00000
00010 1 00000 10001 1 11001
00011 1 00111 10010 2 00000 or 11110
00100 1 00000 10011 2 00111 or 11001
00101 1 00111 10100 2 00000 or 11110
00110 1 00111 10101 2 00111 or 11001
01000 1 00000 10110 1 11110
01001 1 11001 10111 1 00111
01010 2 00000 or 11110 11000 1 11001
01011 2 00111 or 11001 11010 1 11110

6.4 / ERROR CORRECTION 199

01100 2 00000 or 11110 11011 1 11001
01101 2 00111 or 11001 11100 1 11110
01110 1 11110 11101 1 11001
01111 1 00111 11111 1 11110

There are eight cases in which an invalid codeword is at a distance 2 from two
different valid codewords. Thus, if one such invalid codeword is received, an error
in 2 bits could have caused it and the receiver has no way to choose between the
two alternatives. An error is detected but cannot be corrected. However, in every
case in which a single bit error occurs, the resulting codeword is of distance 1
from only one valid codeword and the decision can be made. This code is there-
fore capable of correcting all single-bit errors but cannot correct double bit
errors. Another way to see this is to look at the pairwise distances between valid
codewords:

d100000, 001112 = 3; d100000, 110012 = 3; d100000, 111102 = 4;
d100111, 110012 = 4; d100111, 111102 = 3; d111001, 111102 = 3;

The minimum distance between valid codewords is 3. Therefore, a single bit
error will result in an invalid codeword that is a distance 1 from the original valid
codeword but a distance at least 2 from all other valid codewords. As a result, the
code can always correct a single-bit error. Note that the code also will always
detect a double-bit error.

The preceding example illustrates the essential properties of a block error-
correcting code. An (n, k) block code encodes k data bits into n-bit codewords.
Typically, each valid codeword reproduces the original k data bits and adds to them
1n - k2 check bits to form the n-bit codeword. Thus the design of a block code is
equivalent to the design of a function of the form vc = f1vd2, where vd is a vector of
k data bits and vc is a vector of n codeword bits.
With an (n, k) block code, there are 2 k valid codewords out of a total of 2 n
possible codewords. The ratio of redundant bits to data bits, 1n - k2/k, is called
the redundancy of the code, and the ratio of data bits to total bits, k/n, is called the
code rate. The code rate is a measure of how much additional bandwidth is
required to carry data at the same data rate as without the code. For example, a
code rate of 1/2 requires double the transmission capacity of an uncoded system to
maintain the same data rate. Our example has a code rate of 2/5 and so requires
2.5 times the capacity of an uncoded system. For example, if the data rate input to
the encoder is 1 Mbps, then the output from the encoder must be at a rate of 2.5
Mbps to keep up.
For a code consisting of the codewords w1 , w2 , Á , ws , where s = 2 n, the mini-
mum distance dmin of the code is defined as

[d1wi , wj2]
min
dmin =
i Z j


It can be shown that the following conditions hold. For a given positive integer t,
if a code satisfies dmin Ú 12t + 12, then the code can correct all bit errors up to
and including errors of t bits. If dmin Ú 2t, then all errors … 1t - 12 bits can be
corrected and errors of t bits can be detected but not, in general, corrected. Con-
versely, any code for which all errors of magnitude … t are corrected must satisfy
dmin Ú 12t + 12, and any code for which all errors of magnitude … 1t - 12 are
corrected and all errors of magnitude t are detected must satisfy dmin Ú 2t.
Another way of putting the relationship between dmin and t is to say that the
maximum number of guaranteed correctable errors per codeword satisfies
dmin - 1
t = j k
2
where :x; means the largest integer not to exceed x (e.g., :6.3; = 6). Furthermore,
if we are concerned only with error detection and not error correction, then the
number of errors, t, that can be detected satisfies
t = dmin - 1
To see this, consider that if dmin errors occur, this could change one valid code-
word into another. Any number of errors less than dmin can not result in another
valid codeword.
The design of a block code involves a number of considerations.

1. For given values of n and k, we would like the largest possible value of dmin .
2. The code should be relatively easy to encode and decode, requiring minimal
memory and processing time.
3. We would like the number of extra bits, 1n - k2, to be small, to reduce band-
width.
4. We would like the number of extra bits, 1n - k2, to be large, to reduce error
rate.

Clearly, the last two objectives are in conflict, and tradeoffs must be made.
It is instructive to examine Figure 6.8, based on [LEBO98]. The literature on
error-correcting codes frequently includes graphs of this sort to demonstrate the
effectiveness of various encoding schemes. Recall from Chapter 5 that coding can be
used to reduce the required Eb/N0 value to achieve a given bit error rate.5 The coding
discussed in Chapter 5 has to do with the definition of signal elements to represent
bits. The coding discussed in this chapter also has an effect on Eb/N0 . In Figure 6.8,
the curve on the right is for an uncoded modulation system; the shaded region rep-
resents the area in which improvement can be achieved. In this region, a smaller
BER (bit error rate) is achieved for a given Eb/N0 , and conversely, for a given BER,
a smaller Eb/N0 is required. The other curve is a typical result of a code rate of one-
half (equal number of data and check bits). Note that at an error rate of 10 -6, the
use of coding allows a reduction in Eb/N0 of 2.77 dB. This reduction is referred to as

5
Eb/N0 is the ratio of signal energy per bit to noise power density per Hertz; it is defined and discussed in
Chapter 3.

6.5 / LINE CONFIGURATIONS 201
1

10Ϫ1
Without

Probability of bit error (BER)
coding
10Ϫ2
Channel bit Rate 1/2
error probability coding

10Ϫ3

10Ϫ4

Region of
potential
10Ϫ5 coding gain

10Ϫ6
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
3 dB 2.77 dB
(Eb/N0) (dB)

Figure 6.8 How Coding Improves System Performance

the coding gain, which is defined as the reduction, in decibels, in the required Eb/N0
to achieve a specified BER of an error-correcting coded system compared to an
uncoded system using the same modulation.
It is important to realize that the BER for the second rate 1/2 curve refers to
the rate of uncorrected errors and that the Eb value refers to the energy per data bit.
Because the rate is 1/2, there are two bits on the channel for each data bit, and the
energy per coded bit is half that of the energy per data bit, or a reduction of 3 dB to
a value of 8 dB. If we look at the energy per coded bit for this system, then we see
that the channel bit error rate is about 2.4 * 10-2, or 0.024.
Finally, note that below a certain threshold of Eb/N0, the coding scheme actu-
ally degrades performance. In our example of Figure 6.8, the threshold occurs at
about 5.4 dB. Below the threshold, the extra check bits add overhead to the system
that reduces the energy per data bit causing increased errors. Above the threshold,
the error-correcting power of the code more than compensates for the reduced Eb,
resulting in a coding gain.

6.5 LINE CONFIGURATIONS

Two characteristics that distinguish various data link configurations are topology
and whether the link is half duplex or full duplex.

Topology
The topology of a data link refers to the physical arrangement of stations on a trans-
mission medium. If there are only two stations (e.g., a terminal and a computer or


Terminals
(secondaries)

Host
(primary)

(a) Point-to-point

(b) Multipoint

Figure 6.9 Traditional Computer/Terminal Configurations

two computers), the link is point to point. If there are more than two stations, then it
is a multipoint topology. Traditionally, a multipoint link has been used in the case of
a computer (primary station) and a set of terminals (secondary stations). In today’s
environments, the multipoint topology is found in local area networks.
Traditional multipoint topologies are made possible when the terminals are
only transmitting a fraction of the time. Figure 6.9 illustrates the advantages of the
multipoint configuration. If each terminal has a point-to-point link to its computer,
then the computer must have one I/O port for each terminal. Also there is a sepa-
rate transmission line from the computer to each terminal. In a multipoint configu-
ration, the computer needs only a single I/O port and a single transmission line,
which saves costs.

Full Duplex and Half Duplex
Data exchanges over a transmission line can be classified as full duplex or half
duplex. With half-duplex transmission, only one of two stations on a point-to-point
link may transmit at a time. This mode is also referred to as two-way alternate, sug-
gestive of the fact that two stations must alternate in transmitting. This can be

6.6 / RECOMMENDED READING 203
compared to a one-lane, two-way bridge. This form of transmission is often used for
terminal-to-computer interaction. While a user is entering and transmitting data, the
computer is prevented from sending data to the terminal, which would appear on
the terminal screen and cause confusion.
For full-duplex transmission, two stations can simultaneously send and receive
data from each other. Thus, this mode is known as two-way simultaneous and may be
compared to a two-lane, two-way bridge. For computer-to-computer data exchange,
this form of transmission is more efficient than half-duplex transmission.
With digital signaling, which requires guided transmission, full-duplex opera-
tion usually requires two separate transmission paths (e.g., two twisted pairs), while
half duplex requires only one. For analog signaling, it depends on frequency: If a sta-
tion transmits and receives on the same frequency, it must operate in half-duplex
mode for wireless transmission, although it may operate in full-duplex mode for
guided transmission using two separate transmission lines. If a station transmits on
one frequency and receives on another, it may operate in full-duplex mode for wire-
less transmission and in full-duplex mode with a single line for guided transmission.
It is possible to transmit digital signals simultaneously in both directions on a
single transmission line using a technique called echo cancellation. This is a signal
processing technique whose explanation is beyond the scope of this book.

6.6 RECOMMENDED READING

The classic treatment of error detecting codes and CRC is [PETE61]. [RAMA88] is an excel-
lent tutorial on CRC.
[STAL05] discusses most of the widely used error-correcting codes. [ADAM91] pro-
vides comprehensive treatment of error-correcting codes. [SKLA01] contains a clear, well-
written section on the subject. Two useful survey articles are [BERL87] and [BHAR83]. A
quite readable theoretical and mathematical treatment of error-correcting codes is [ASH90].
[FREE98] provides good coverage of many physical layer interface standards.

ADAM91 Adamek, J. Foundations of Coding. New York: Wiley, 1991.
ASH90 Ash, R. Information Theory. New York: Dover, 1990.
BERL87 Berlekamp, E.; Peile, R.; and Pope, S. “The Application of Error Control to
Communications.” IEEE Communications Magazine, April 1987.
BHAR83 Bhargava, V. “Forward Error Correction Schemes for Digital Communica-
tions.” IEEE Communications Magazine, January 1983.
FREE98 Freeman, R. Telecommunication Transmission Handbook. New York: Wiley,
1998.
PETE61 Peterson, W., and Brown, D. “Cyclic Codes for Error Detection.” Proceedings
of the IEEE, January 1961.
RAMA88 Ramabadran, T., and Gaitonde, S. “A Tutorial on CRC Computations.” IEEE
Micro, August 1988.
SKLA01 Sklar, B. Digital Communications: Fundamentals and Applications. Upper Sad-
dle River, NJ: Prentice Hall, 2001.
STAL05 Stallings, W. Wireless Communications and Networks, Second Edition. Upper
Saddle River, NJ: Prentice Hall, 2005.


6.7 KEY TERMS, REVIEW QUESTIONS, AND PROBLEMS

Key Terms

asynchronous transmission error-detecting code interchange circuits
codeword forward error correction (FEC) Integrated Services Digital
cyclic code frame Network (ISDN)
cyclic redundancy check (CRC) frame check sequence (FCS) modem
EIA-232 full duplex parity bit
error correction half duplex parity check
error-correcting code (ECC) Hamming code point-to-point
error detection Hamming distance synchronous transmission

Review Questions
6.1. How is the transmission of a single character differentiated from the transmission of
the next character in asynchronous transmission?
6.2. What is a major disadvantage of asynchronous transmission?
6.3. How is synchronization provided for synchronous transmission?
6.4. What is a parity bit?
6.5. What is the CRC?
6.6. Why would you expect a CRC to detect more errors than a parity bit?
6.7. List three different ways in which the CRC algorithm can be described.
6.8. Is it possible to design an ECC that will correct some double bit errors but not all
double bit errors? Why or why not?
6.9. In an (n, k) block ECC, what do n and k represent?

Problems
6.1 Suppose a file of 10,000 bytes is to be sent over a line at 2400 bps.
a. Calculate the overhead in bits and time in using asynchronous communication.
Assume one start bit and a stop element of length one bit, and 8 bits to send the
byte itself for each character. The 8-bit character consists of all data bits, with no
parity bit.
b. Calculate the overhead in bits and time using synchronous communication.
Assume that the data are sent in frames. Each frame consists of 1000 characters ϭ
8000 bits and an overhead of 48 control bits per frame.
c. What would the answers to parts (a) and (b) be for a file of 100,000 characters?
d. What would the answers to parts (a) and (b) be for the original file of 10,000 char-
acters except at a data rate of 9600 bps?
6.2 A data source produces 7-bit IRA characters. Derive an expression of the maximum
effective data rate (rate of IRA data bits) over an x-bps line for the following:
a. Asynchronous transmission, with a 1.5-unit stop element and a parity bit.
b. Synchronous transmission, with a frame consisting of 48 control bits and 128 infor-
mation bits. The information field contains 8-bit (parity included) IRA characters.
c. Same as part (b), except that the information field is 1024 bits.
6.3 Demonstrate by example (write down a few dozen arbitrary bit patterns; assume one
start bit and a stop element of length one bit) that a receiver that suffers a framing
error on asynchronous transmission will eventually become realigned.

6.7 / KEY TERMS, REVIEW QUESTIONS, AND PROBLEMS 205
6.4 Suppose that a sender and receiver use asynchronous transmission and agree
not to use any stop elements. Could this work? If so, explain any necessary conditions.
6.5 An asynchronous transmission scheme uses 8 data bits, an even parity bit, and a stop
element of length 2 bits. What percentage of clock inaccuracy can be tolerated at the
receiver with respect to the framing error? Assume that the bit samples are taken at
the middle of the clock period. Also assume that at the beginning of the start bit the
clock and incoming bits are in phase.
6.6 Suppose that a synchronous serial data transmission is clocked by two clocks (one at
the sender and one at the receiver) that each have a drift of 1 minute in one year.
How long a sequence of bits can be sent before possible clock drift could cause a
problem? Assume that a bit waveform will be good if it is sampled within 40% of its
center and that the sender and receiver are resynchronized at the beginning of each
frame. Note that the transmission rate is not a factor, as both the bit period and the
absolute timing error decrease proportionately at higher transmission rates.
6.7 Would you expect that the inclusion of a parity bit with each character would change
the probability of receiving a correct message?
6.8 Two communicating devices are using a single-bit even parity check for error detec-
tion. The transmitter sends the byte 10101010 and, because of channel noise, the
receiver gets the byte 10011010. Will the receiver detect the error? Why or why not?
6.9 What is the purpose of using modulo 2 arithmetic rather than binary arithmetic in
computing an FCS?
6.10 Consider a frame consisting of two characters of four bits each. Assume that the prob-
ability of bit error is 10-3 and that it is independent for each bit.
a. What is the probability that the received frame contains at least one error?
b. Now add a parity bit to each character. What is the probability?
6.11 Using the CRC-CCITT polynomial, generate the 16-bit CRC code for a message con-
sisting of a 1 followed by 15 0s.
a. Use long division.
b. Use the shift register mechanism shown in Figure 6.6.
6.12 Explain in words why the shift register implementation of CRC will result in all 0s at
the receiver if there are no errors. Demonstrate by example.
6.13 For P = 110011 and M = 11100011, find the CRC.
6.14 A CRC is constructed to generate a 4-bit FCS for an 11-bit message. The generator
polynomial is X4 + X3 + 1.
a. Draw the shift register circuit that would perform this task (see Figure 6.6).
b. Encode the data bit sequence 10011011100 (leftmost bit is the least significant)
using the generator polynomial and give the codeword.
c. Now assume that bit 7 (counting from the LSB) in the codeword is in error and
show that the detection algorithm detects the error.
6.15 a. In a CRC error-detecting scheme, choose P1x2 = x 4 + x + 1. Encode the bits
10010011011.
b. Suppose the channel introduces an error pattern 100010000000000 (i.e., a flip
from 1 to 0 or from 0 to 1 in position 1 and 5). What is received? Can the error be
detected?
c. Repeat part (b) with error pattern 100110000000000.
6.16 A modified CRC procedure is commonly used in communications standards. It is
defined as follows:

X16D1X2 + XkL1X2 R1X2
= Q +
P1X2 P1X2
FCS = L1X2 + R1X2


where
L1X2 = X15 + X14 + X13 + Á + X + 1

and k is the number of bits being checked (address, control, and information fields).
a. Describe in words the effect of this procedure.
b. Explain the potential benefits.
c. Show a shift register implementation for P1X2 = X16 + X12 + X5 + 1.
6.17 Calculate the Hamming pairwise distances among the following codewords:
a. 00000, 10101, 01010
b. 000000, 010101, 101010, 110110
6.18 Section 6.4 discusses block error-correcting codes that make a decision on the
basis of minimum distance. That is, given a code consisting of s equally likely
codewords of length n, for each received sequence v, the receiver selects the
codeword w for which the distance d(w, v) is a minimum. We would like to prove
that this scheme is “ideal” in the sense that the receiver always selects the code-
word for which the probability of w given v, p1w ƒ v2, is a maximum. Because all
codewords are assumed equally likely, the codeword that maximizes p1w ƒ v2 is the
same as the codeword that maximizes p1v ƒ w2.
a. In order that w be received as v, there must be exactly d(w, v) errors in transmis-
sion, and these errors must occur in those bits where w and v disagree. Let b be
the probability that a given bit is transmitted incorrectly and n be the length of a
codeword. Write an expression for p1v ƒ w2 as a function of b, d(w, v), and n. Hint:
The number of bits in error is d(w, v) and the number of bits not in error is
n - d1w, v2.
b. Now compare p1v ƒ w12 and p1v ƒ w22 for two different codewords w1 and w2 by cal-
culating p1v ƒ w12/p1v ƒ w22.
c. Assume that 0 6 b 6 0.5 and show that p1v ƒ w12 7 p1v ƒ w22 if and only if
d1v, w12 6 d1v, w22. This proves that the codeword w that gives the largest value
of p1v ƒ w2 is that word whose distance from v is a minimum.
6.19 Section 6.4 states that for a given positive integer t, if a code satisfies dmin Ú 2t + 1,
then the code can correct all bit errors up to and including errors of t bits. Prove this
assertion. Hint: Start by observing that for a codeword w to be decoded as another
codeword w œ , the received sequence must be at least as close to w œ as to w.
Note: The remaining problems concern material in Appendix G.
6.20 Draw a timing diagram showing the state of all EIA-232 leads between two DTE-
DCE pairs during the course of a data call on the switched telephone network.
6.21 Explain the operation of each null modem connection in Figure G.5.
6.22 For the V.24/EIA-232 Remote Loopback circuit to function properly, what circuits
must be logically connected?

DATA LINK CONTROL PROTOCOLS
CHAPTER
7
7.1 Flow Control

7.2 Error Control

7.3 High-Level Data Link Control (HDLC)

7.4 Recommended Reading

7.5 Key Terms, Review Questions, and Problems

Appendix 7A Performance Issues

207

208 CHAPTER 7 / DATA LINK CONTROL PROTOCOLS

“Great and enlightened one,” said Ten-teh, as soon as his stupor was lifted, “has this
person delivered his message competently, for his mind was still a seared vision of
snow and sand and perchance his tongue has stumbled?”
“Bend your ears to the wall,” replied the Emperor, “and be assured.”

—Kai Lung’s Golden Hours, Earnest Bramah

KEY POINTS
• Because of the possibility of transmission errors, and because the
receiver of data may need to regulate the rate at which data arrive, syn-
chronization and interfacing techniques are insufficient by themselves. It
is necessary to impose a layer of control in each communicating device
that provides functions such as flow control, error detection, and error
control. This layer of control is known as a data link control protocol.
• Flow control enables a receiver to regulate the flow of data from a
sender so that the receiver’s buffers do not overflow.
• In a data link control protocol, error control is achieved by retrans-
mission of damaged frames that have not been acknowledged or for
which the other side requests a retransmission.
• High-level data link control (HDLC) is a widely used data link con-
trol protocol. It contains virtually all of the features found in other
data link control protocols.

Our discussion so far has concerned sending signals over a transmission link. For
effective digital data communications, much more is needed to control and man-
age the exchange. In this chapter, we shift our emphasis to that of sending data
over a data communications link. To achieve the necessary control, a layer of logic
is added above the physical layer discussed in Chapter 6; this logic is referred to as
data link control or a data link control protocol. When a data link control protocol
is used, the transmission medium between systems is referred to as a data link.
To see the need for data link control, we list some of the requirements and
objectives for effective data communication between two directly connected
transmitting-receiving stations:
• Frame synchronization: Data are sent in blocks called frames. The beginning
and end of each frame must be recognizable.We briefly introduced this topic
with the discussion of synchronous frames (Figure 6.2).
• Flow control: The sending station must not send frames at a rate faster
than the receiving station can absorb them.
• Error control: Bit errors introduced by the transmission system should be
corrected.
• Addressing: On a shared link, such as a local area network (LAN), the
identity of the two stations involved in a transmission must be specified.

Ch 6 data and computer communicationwilliam stallings

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